Question

one gram of charcoal having surface area 3.02 into 10 to the power 2 m square is mixed in hundred ml of 0.5 m Acetic Acid solution after few times concentration of solution because 0.49 M what will be surface area occupied one mole of Acetic Acid

Answer 1

Number of moles of acetic in 100 ml before adding charcoal = 0.05

Number of moles of acetic acid in 100 ml after adding charcoal = 0.049

Number of moles of acetic acid adsorbed on the surface of charcoal = 0.001

Number of molecules of acetic acid adsorbed on the surface of charcoal = 0.001 * 6.02 * 1023 = 6.02 * 1020

Surface area of charcoal = 3.01 * 102 m2(given)

Area occupied by single acetic acid molecule on the surface

of charcoal 3.01 *102/6.02 * 1020 = 5 *10-19m2

Answer 2

Answer:5 *10-19m2

Explanation:

given,

molarity of acetic acid=0.5M

volume=100ml ie,0.1L

no of moles=molarity X volume

Therefore no of moles =0.05

similarly moles left after adsorption=0.49M X 0.1L =0.049

hence no of moles of adsorbed=0.05-0.049=0.001 mole

0.001 mole corresponds to 6.02 X 10^20 molecules

surface area occupied by each molecule=total surface area of charcoal/number of molecules of acetic acid

=3.02 X 10^2/6.02 X 10^20

=5 X 10^-19 m^2

Click to let others know, how helpful is it

4.0

69 votes

THANKS 151

Report

Budaniashray74p8b5n1Ambitious

Answer: 5×10^-19 m2

Explanation:

Number of moles of acetic in 100 ml before adding charcoal = 0.05

Number of moles of acetic acid in 100 ml after adding charcoal = 0.049

Number of moles of acetic acid adsorbed on the surface of charcoal = 0.001

Number of molecules of acetic acid adsorbed on the surface of charcoal = 0.001 * 6.02 * 1023 = 6.02 * 1020

Surface area of charcoal = 3.01 * 102 m2(given)

Area occupied by single acetic acid molecule on the surface

of charcoal 3.01 *102/6.02 * 1020 = 5 *10-19m2